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Basic proof of units in a ring with identity.

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Let R be a ring with identity, and a,b are elements in R. If ab is a unit, and neither a nor b is a zero divisor, prove a and b are units.


    2. Relevant equations

    If ab is a unit then (ab)c=1=c(ab) for some c in R.


    3. The attempt at a solution
    Assume both a and b are not zero divisors, and denote the identity element of R as 1.

    Since ab is a unit in R, there exists some c such that (ab)c = 1.
    (ab)c=1 [itex]\Rightarrow[/itex] (ca)b = 1 [itex]\Rightarrow[/itex] b is a unit.

    Similarly, (ab)c= 1 [itex]\Rightarrow[/itex] a(bc) [itex]\Rightarrow[/itex] a is a unit.

    Does the above sufficiently prove the claim in the problem statement? I feel like I'm missing something.

    Thanks!
     
  2. jcsd
  3. Feb 28, 2013 #2

    Dick

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    Yes, you should feel uncomfortable with that. You didn't use that a and b weren't zero divisors. If (ab)c=1 then you can regroup that to a(bc)=1. That doesn't prove a is a unit. You also need to show (bc)a=1. Don't assume the ring is commutative.
     
  4. Mar 1, 2013 #3
    Ah thank you, that was a foolish oversight, they wouldn't stipulate a and b weren't zero divisors if it wasn't necessary...

    So if we let a, b and c be elements of R such that ab = ac, then:

    ab-ac=0R = a(b-c) = 0R, and
    a is not a zero divisor [itex]\Rightarrow[/itex] (b - c) = 0R [itex]\Rightarrow[/itex] (b-c) = 0R [itex]\Rightarrow[/itex] b = c

    Does that allow me to claim the following?

    Multiplying a(bc) = 1 by a to give a(bc)a = 1a , (1a=a1 by definition) then, a(bc)a = a1 and by cancellation, (bc)a = 1 [itex]\Rightarrow[/itex] a is a unit.
     
  5. Mar 1, 2013 #4

    Dick

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    Yes, that's the idea.
     
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